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    Re: Handling scatter
    From: Andr�s Ruiz
    Date: 2008 Jun 16, 12:22 +0200

    This paper, How to Average Celestial Sights for Optimum Accuracy, by David Burch can be found at:
    -----Mensaje original-----
    De: NavList@fer3.com [mailto:NavList@fer3.com] En nombre de George Huxtable
    Enviado el: s�bado, 14 de junio de 2008 1:46
    Para: NavList@fer3.com
    Asunto: [NavList 5430] Handling scatter. was [NavList 5421] Re: Exercise #12 Daylight Sun/Moon Fix
    Mike Burkes wrote (reassembled)-
    >  I noticed a number of members averaged
    > the entire moon set but upon my graphing the set it becomes readily
    > apparent sites 21-01-40 and 21-03-22 are rejected therefore the line
    > of best fit falls nicely thru the remaining 6 sites and solving no 3,
    > the 21-00-48 site, yields an agreeable solution  and a run-on sentence
    Response from George.
    I've made Jeremy's 8 Moon observations, being discussed here,  available as 
    an Excel attachment to illustrate the points involved.
    Mike Burkes has his own way of doing these things, but I suggest that the 
    procedure he describes should not be copied. It may be of interest to the 
    list to start a bit of discussion about methods of handling scatter, to 
    discover the attitude of others, and Mike's message provides a good starting 
    example. I will present my own views, which I like to think of as 
    "scientific", to provoke some comment. I will go into this particular 
    sight-reduction in some detail, not because it's an important matter in 
    itself, but because it provides such an example of different approaches. I 
    hope Mike isn't too sensitive to being used as an example in this way, and 
    hope he'll argue back if he thinks fit.
    Jeremy has explained that the Moon limb was unsharp because the contrast 
    with the bright sky background was very low, and therefore he was not 
    confident about the precision of those lunar altitudes. Presumably, that 
    uncertainty applied equally to each of the 8 observations, which show an 
    unexpected amount of scatter.
    Mike wrote "but upon my graphing the set it becomes readily apparent sites 
    21-01-40 and 21-03-22 are rejected ...". Rejected by Mike, but on what 
    grounds? In my view, the only valid reason for rejecting some sight from a 
    set, and accepting others, is if it is so far out of line that it must have 
    been the result of a blunder. If, on the other hand, it's the result of 
    scatter due to difficult observing conditions, then one such observation is 
    as good as any other, and the way to deal with the scatter is simply to 
    average the lot, giving equal weight to every point. Because outliers are 
    fewer in numbers than the main group, their contribution to the averaged 
    result is correspondingly limited.
    However, if I were in a generous mood, I might be persuaded to go along with 
    Mike's rejection of the very highest observation, made at 21:01:40, as being 
    particularly far out of line with the rest, and perhaps plausibly the result 
    of some sort of blunder; in which case I might reluctantly accept leaving 
    that point out and averaging the rest.  But what justification can he then 
    offer for rejecting also the next-highest, at 23:01:22, while accepting the 
    lowest, at 21:05:08? One is no further out-of-line with the rest of the 
    group than is the other.
    But in the end Mike rejects much more than than. He looks at the six points 
    that remain after discarding the highest two, and decides that on some basis 
    he can draw a line through them, a line which happens to pass through the 
    sight taken at 21:00:48. So he then discards everything else and works on 
    the basis of that one point only. This process may have saved him a bit of 
    arithmetic, in avoiding that averaging, but I suggest that a methodical 
    averaging process would have squeezed the most out of the available 
    As I side-issue, I've come across a procedure, adopted by some, when faced 
    with the problem of reducing the scatter in a large number of observations 
    of what ought to be the same thing, by doing the following-
    Reduce the range by rejecting both the highest and the lowest observation in 
    the set, as a pair, before averaging the rest. Indeed, if there are 
    sufficient observations, weed out the next remaing pair in the same way, and 
    so on. In the extreme, this shifts the resulting answer towards the median 
    of the set, rather than the mean. If the distribution is truly Gaussian, it 
    does nothing to reduce the scatter, compared with simple averaging of the 
    whole set; indeed, it increases it, though only slightly. The only virtue of 
    this approach appears when the distribution is not truly Gaussian; if there 
    are more outliers, further out from the average than ordinary statistics 
    would predict (blunders, perhaps). Then a procedure such as this can provide 
    a methodical way of weeding out such blunders.
    contact George Huxtable at george@huxtable.u-net.com
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    Navigation List archive: www.fer3.com/arc
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